In `S_(N)1` reaction of `(S)-2` bromobutane with water, two substitution products are formed one has the same relative configuration as the reactant and the other has the inverted configuration. This is in most `S_(N)1` reactions, the racemic product is non-50:50 mixture i.e, racemization takes place the inverted product is more than 50% . The formation of different composition of the product is due to following steps involved in `S_(N)` reactions.
`R-X " " to R^(+) X^(-) to R^(+) " " "solvent" X^(-) " " to R^(+)+X^(-)`
`{:("Undissociated Molecule", "Intimate ion pair", "solvent-separated ion pair", "dissociated ions"),(I,II,III,IV):}`
In the above reaction the non-50 :50 racemic mixture contains more amount of

Options

A. original configuration
B. inverted configuration
C. either of the two
D. none of the above

Answer

To solve the question regarding the SN1 reaction of (S)-2-bromobutane with water and the formation of substitution products, we can break down the process step by step.

### Step-by-Step Solution:

1. **Understanding the Reactant**:
- The reactant is (S)-2-bromobutane, which has a specific stereochemistry. The presence of the bromine atom (Br) at the second carbon creates a chiral center.

**Hint**: Identify the stereochemistry of the reactant and the significance of the chiral center.

2. **Formation of Carbocation**:
- In an SN1 reaction, the first step involves the departure of the leaving group (Br) to form a carbocation. This carbocation is planar and can be attacked from either side by the nucleophile (water in this case).

**Hint**: Remember that the formation of a carbocation is a key step in SN1 reactions, leading to a planar intermediate.

3. **Nucleophilic Attack**:
- The nucleophile (water) can attack the carbocation from either side. This results in two possible products: one with the same configuration as the reactant and one with the inverted configuration.

**Hint**: Consider how the nucleophile can approach the planar carbocation from different angles, leading to different configurations.

4. **Racemization and Product Distribution**:
- Although both configurations are formed, the product distribution is not 50:50. Due to steric hindrance from the bromine atom, the nucleophile is more likely to attack from the side opposite to the bromine, leading to a predominance of the inverted product.

**Hint**: Analyze the steric effects and how they influence the likelihood of nucleophilic attack on the carbocation.

5. **Conclusion**:
- In the non-50:50 racemic mixture, the inverted configuration will be present in greater amounts than the same configuration. This is because the steric hindrance from the bromine atom makes the backside attack (leading to inversion) more favorable.

**Hint**: Summarize the key factors that lead to the predominance of one product over the other in the racemic mixture.

### Final Answer:
In the non-50:50 racemic mixture formed from the SN1 reaction of (S)-2-bromobutane with water, the inverted configuration will be present in greater amounts than the same configuration.